Subject: Chemistry — Ionic Equilibrium
Question: The solubility of a salt of weak acid (AB) at pH 3 is
$Y \times 10^{- 3}\ mol\ L^{- 1}$. The value of $Y$ is
\_\_\_\_\_\_\_\_\_\_.
(Given that the value of solubility product of
$AB\ \left( K_{sp} \right)$ $= 2 \times 10^{- 10}$ and the value of
ionization constant of $HB\ \left( K_{a} \right) = 1 \times 10^{- 8}$)
Answer: 4.47
Solution: $AB \rightleftharpoons \ A^{+}\ \ + \ B^{-}$
$x$ $x$ $x - y$
$2 \times 10^{10} = x(x - y)$ ... (1)
$B^{-}\ \ \ \ + \ \ H^{+} \rightleftharpoons HB$
$x - y$ $10^{- 3}$ $y$
$$10^{8} = \frac{y}{(x - y)10^{- 3}}$$
$$\frac{y}{x - y} = 10^{5}$$
$x - y = 10^{- 5}y$ ... (2)
From (1) & (2)
$$2 \times 10^{- 10} = x^{2} - 2 \times 10^{- 5}$$
$$x^{2} = 2 \times 10^{- 5}$$
$$x = \sqrt{20} \times 10^{- 3} = 4.47 \times 10^{- 3}$$